# What is dy/dx for y - ln(1/y) + ln(y) = ln(x).

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### 2 Answers

y- ln (1/y) + ln y = ln x.

First we will rewrite ln (1/y) = ln y^-1 = - ln y.

==> y + ln y + ln y = ln x.

==> y + 2*ln y = ln x.

Now we will differentiate with respect to x.

==> (y)' + (2ln y)' = (ln x) '

==> y' + 2*1/y * y' = 1/x

==> y' + (2/y) y' = 1/x

Now we will factor y':

==> y' ( 1+ 2/y) = 1/x

==> y' ( y+2)/y = 1/x

Now we will multiply both sides by y/(y+2)

==> y' = 1/x * ( y/(y+2)

==> y' = y/x(y+2)

**Then, dy/dx = y/x(y+2)**

We have y - ln ( 1/y) + ln y = ln x

use the relation that log a + log b = log a*b

=> y - ln [( 1/y)(1/y)] = ln x

=> y - ln [ 1/y^2] = ln x

=> dy/dx - y^2*(-2)*(1/y^3)*dy/dx = 1/x

=> dy/dx [ 1 + 2* y^2 / y^3] = 1/x

=> dy/dx [ 1 + 2/y]= 1/x

=> dy/dx = 1/[x*(1 + 2/y)]

**The required value of dy/dx = 1/[x*(1 + 2/y)]**